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How many ways are there to arrange the letters of the word GARDEN with the vowels in alphabetical order?

I tried solving it, but I'm not even getting a clue.

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    $\begingroup$ Why did you mark the $480$ answer? (I am pretty sure it's wrong, it should be $360$). $\endgroup$ – Leo163 Aug 14 '17 at 9:34
  • $\begingroup$ Am I making a mistake, or are the options for Q12 all wrong? I think the answer should be $5!\times6!/2!$. (There are $5!$ ways to order the men, up to cyclic shifts, then $6$ choices for which two men the first woman sits between, $5$ for the next and so on.) $\endgroup$ – Especially Lime Aug 14 '17 at 9:54
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    $\begingroup$ Please type your questions rather than posting an image. Images cannot be searched. $\endgroup$ – N. F. Taussig Aug 14 '17 at 9:56
  • $\begingroup$ @EspeciallyLime Woah, going out of syllabus there, man! :D $\endgroup$ – ab123 Aug 14 '17 at 10:16
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So the vowels in "GARDEN" are 'A' and 'E'. Total permutations of the word GARDEN are $6! = 720.$ In half of them, A will occur before E and in the other half of the permutaions, E will occur before A. This is obvious as "A before E" and "E before A" are equally likely and exclusive(nothing other than these two can happen) events.

So both must have probability $1/2$. Hence in half of the words, vowels are in alphabetical order. Hence, answer is $720/2 = 360$.

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  • $\begingroup$ @bof Oh yes, that was a mistake. Thanks. I edited it to equally likely and exclusive events. "A before E" and "E before A" are not independent as if one occurs, probability of the other occuring becomes 0. $\endgroup$ – ab123 Aug 14 '17 at 10:24
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There are $5+4+3+2+1 = 15$ ways to arrange the vowels $A$, and $E$ in alphabetical order. After that, you have $4! = 24$ ways to arrange the remaining $4$ consonants . Thus the total number of words formed this way is: $15\cdot 24 = 360$

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You can start with the combination $AEgrdn$. The consonants can be arranged in $4!=24$ ways. Now the letter $E$ can be put on the positions $2,3,4,5,6$. Thus in total we have $5\cdot 24=120$ with A at position 1 and E behind.

Next $A$ is put on position 2: $gAErdn$. Again the consonants can be arranged in $24$ ways. And $E$ can be put on the positions $3,4,5,6$. Therefore in total we have $4\cdot 24=96$ with $A$ at position $2$ and $E$ behind.

Proceed in this manner till $A$ at position $5$ and $E$ at postion $6$.

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